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The domain of a cartesian function from parametric equations

  1. Jun 10, 2008 #1
    x = 2cot t

    y = (sin t)^2

    t is greater than 0 but less than or equal to pi/2


    The cartesian can be found using trig identities to be:

    y = 8/ (4+ x^2)


    What would be the range of the cartesian equation? I think it would be x is greater than or equal to 0, since when t = pi/2, x = 0, and as t tends to 0, x tends to infinity.

    Am I correct?

    Thank you.
     
  2. jcsd
  3. Jun 10, 2008 #2

    HallsofIvy

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    Your title says "domain" but in the body of your message you say "range". Which is it?

    If you are thinking of the "cartesian equation" as y a function of x, then the domain is the set of all possible x values which is the set of all vaues of cot(t) for t between 0 and pi/2 and the range is the set of all y values which is the set of all values of (sin t)^2 for t between 0 and pi/2.
     
  4. Jun 11, 2008 #3
    Sorry for the confusion! I meant the domain! Would my answer therefore be correct?

    Thanks
     
  5. Jun 11, 2008 #4

    HallsofIvy

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